Is Fg the Only Force to Consider for Ft4 in a Pulley System?

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SUMMARY

In a frictionless pulley system, the tension forces Ft1 and Ft2 are equal, and the gravitational force Fg is a critical factor when calculating Ft4. When lifting an object, such as a piano, the force applied must be considered for Ft3, which is equal to the mass times gravitational acceleration (m*g). If the piano moves at a constant speed, Ft4 equals the weight of the piano, necessitating additional force to accelerate it upward. The configuration of the pulley system, specifically the number of strands supporting the load, directly influences the tension in each strand.

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  • Understanding of Newton's laws of motion
  • Knowledge of basic physics concepts related to forces and tension
  • Familiarity with pulley systems and their mechanics
  • Ability to calculate gravitational force (Fg = m*g)
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  • Study the principles of tension in multi-strand pulley systems
  • Learn about the effects of friction in pulley systems and how to account for it
  • Explore the dynamics of objects in motion, particularly under constant velocity
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I have some questions about the following problem:

I know that Ft1 and Ft2 are equal. Is Fg the only force that needs to be taken into account for Ft4? Does the force being applied to lift the piano need to be taken into account for Ft3? If not, is it just m*g?

http://img96.imageshack.us/img96/8798/782tm.jpg
 
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Assuming that the pulleys are frictionless, the tension is the same throughout the rope (frictionless pulleys can be found only in textbook problems).

Also assuming that the piano is moving at a constant speed, F_T4 is equal to the weight of the piano. More force here is necessary to accelerate the piano upward.

Look at the top pully: how many strands are pulling it down? How much tension in each strand? ("strand" = "rope segment")
 

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